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2 years ago

When the sun and moon work together ____________ are formed on Earth.

Physics

1 answer:

Paha777 [63]2 years ago

7 0

Answer:

C. Spring Tides is the answer.

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b) The distance of the red supergiant Betelgeuse is approximately 427 light years. If it were to explode as a supernova, it woul

Nat2105 [25]

Answer:

b) Betelgeuse would be $\approx 1.43 \cdot 10^{6}$ times brighter than Sirius

c) Since Betelgeuse brightness from Earth compared to the Sun is $\approx 1.37 \cdot 10^{-5} }$ the statement saying that it would be like a second Sun is incorrect

Explanation:

The start brightness is related to it luminosity thought the following equation:

$B = \displaystyle{\frac{L}{4\pi d^2}}$ (1)

where $B$ is the brightness, $L$ is the star luminosity and $d$ , the distance from the star to the point where the brightness is calculated (measured). Thus:

b) $B_{Betelgeuse} = \displaystyle{\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2}}$ and $B_{Sirius} = \displaystyle{\frac{26L_{Sun}}{4\pi (26\ ly)^2}}$ where $L_{Sun}$ is the Sun luminosity ( $3.9 x 10^{26} W$ ) but we don't need to know this value for solving the problem. $ly$ is light years.

Finding the ratio between the two brightness we get:

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sirius}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (26\ ly)^2}{26L_{Sun}} \approx 1.43 \cdot 10^{6} }$

c) we can do the same as in b) but we need to know the distance from the Sun to the Earth, which is $1.581 \cdot 10^{-5}\ ly$ . Then

$\displaystyle{\frac{B_{Betelgeuse}}{B_{Sun}}=\frac{10^{10}L_{Sun}}{4\pi (427\ ly)^2} \times \frac{4\pi (1.581\cdot 10^{-5}\ ly)^2}{1\ L_{Sun}} \approx 1.37 \cdot 10^{-5} }$

Notice that since the star luminosities are given with respect to the Sun luminosity we don't need to use any value a simple states the Sun luminosity as the unit, i.e 1. From this result, it is clear that when Betelgeuse explodes it won't be like having a second Sun, it brightness will be 5 orders of magnitude smaller that our Sun brightness.

4 0

3 years ago

A 0.144 kg baseball is moving towards home plate with a speed of 43.0 m/s when it is bunted. the bat exerts an average force of

Ganezh [65]

As per impulse momentum theorem we know that

$F\Delta t = m(v_f - v_i)$

now here we will have

$F = 6.50 \times 10^3 N$

t = 1.30 ms

m = 0.144 kg

$v_i = -43 m/s$

now we need to find final speed using above formula

$(6.50 \times 10^3)(1.30 \times 10^{-3}) = 0.144 ( v_f - (-43))$

$v_f = 15.7 m/s$

so final speed is given as above

6 0

3 years ago

A single strain gage has a nominal resistance of 120 ohm. For a quarter bridge with 120 ohm fixed resistors, strain gauge factor

natita [175]

Answer:

Output voltage is 1.507 mV

Solution:

As per the question:

Nominal resistance, R = $120\Omega$

Fixed resistance, R = $120\Omega$

Gauge Factor, G.F = 2.01

Supply Voltage, $V_{s} = 3\ V$

Strain, $\epsilon = 1000\times 10^{-6}\ strain$

Now,

To calculate the output voltage, $V_{o}$ :

WE know that strain is given by:

$\epsilon = \frac{(R + R')^{2}V_{o}}{RR'V_{s}\times G.F}$

Thus

$V_{o} = \frac{RR'V_{s}\epsilon \times G.F}{(R + R')^{2}}$

Now, substituting the suitable values in the above eqn:

$V_{o} = \frac{120\times 120\times 3\times 1000\times 10^{-6}\times 2.01}{(120 + 120)^{2}}$

$V_{o} = 1.507\ mV$

6 0

3 years ago

A submarine is at an ocean depth of 3.00 km.

Alexxx [7]

Answer:

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7 0

3 years ago

If an object is projected upward from ground level with an initial velocity of 144144 ft per sec, then its height in feet after

Liula [17]

Answer:

4.5 s, 324 ft

Explanation:

The object is projected upward with an initial velocity of

$v_0 = 144 ft/s$

The equation that describes its height at time t is

$s(t) = -16t^2 + 144 t$ (1)

where t, the time, is measured in seconds.

In order to find the time it takes for the object to reach the maximum height, we must find an expression for its velocity at time t, which can be found by calculating the derivative of the position, s(t):

$v(t) = s'(t) = -32t +144$ (2)

At the maximum heigth, the vertical velocity is zero:

v(t) = 0

Substituting into the equation above, we find the corresponding time at which the object reaches the maximum height:

$0=-32t+144\\t=\frac{144}{32}=4.5 s$

And by substituting this value into eq.(1), we also find the maximum height:

$s(t) = -16(4.5)^2+144(4.5)=324 ft$

3 0

3 years ago